Integration. Section 8: Using partial fractions in integration

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1 Integration Section 8: Using partial fractions in integration Notes and Eamples These notes contain subsections on Using partial fractions in integration Putting all the integration techniques together Using partial fractions in integration The technique of splitting a fraction into partial fractions can be used to integrate some functions involving fractions, as the following two eamples illustrate. Eample Find d Solution The integrand splits into two partial fractions, like this: A B ( )( ) Multiplying through by ( + )( ): = A( ) + B( + ) Substituting = : B B Substituting = : A A So ln c Eample Evaluate d, giving your answer eactly. MEI, 7/0/07 /7

2 Solution A B C ( ) Multiplying through by ²( + ): + = A( + ) + B( + ) + C Substituting = : C C Substituting = 0: A A Equating coefficients of : 0 B C B So ln ln( ) ln ln ln ln ln ln ln ln You may also like to look at the Integrating algebraic fractions video. Putting all the integration techniques together The weight of integration facts and techniques can be a little daunting! Choosing the right technique to solve each problem is not always obvious. However, there are two basic ways to become confident at integration: Know your standard derivative and integral results Wherever possible, do integrals by inspection. Here is a summary of the integration techniques so far, and when to use them: Standard integrals Learn these! Function Integral Function Integral n n sin cos c c for n - n cos sin c ln c e e c sec tan c MEI, 7/0/07 /7

3 Simple variations of these should also be learned: Function e k sin k cos k Integral e k c k cos k c k sin k c k These results can be found using a substitution u = k, but this is usually done in your head. This technique is sometimes called the reverse chain rule. Eample Find the following integrals: sin e Solution sin cos c e e c Integration by inspection Other variations of the standard functions can be integrated by guessing what the answer is and differentiating this. If the result is within a constant multiple of the integral, adjust accordingly. Functions which can be integrated by inspection can be recognised. An integral of the form df g[f ( )] (i.e. the proct of a function of f() and the derivative of f()) can be integrated by inspection. The integral with be of the form of g( ) A particular type of integral which you should recognise is part (iii) of Eample ). f ( ) d ln f( ) c (see f( ) Eample Find using integration by inspection: sin cos d (iii) d MEI, 7/0/07 /7

4 Solution The integral is of the form The integrand is a proct of (a function of + ²), and (a multiple of the derivative of + ²). (iii) d ( ) ( ) So ( ) which is times the integral. d ( ) c sin cos d The integral is of the form cos. d (cos ) cos sin sin cos which is times the integral. So sin cos cos c d The integral is of the form ln. d ln( ) which is times the integral. So d ln( ) c The integrand is a proct of cos (a function of cos ) and sin, (a multiple of the derivative of cos ). The integrand is of the form f ( ) f ( ), where f ( ) With practice you may find that you can do the differentiation in your head and make the necessary adjustment. Integration by substitution Many of the simpler eamples can be done by inspection, as can be seen by the following eample, in which eamples, and (iii) are as in Eample. Eample 5 Find using integration by substitution: (iii) d (iv) sin cos d 0 MEI, 7/0/07 /7

5 Solution u u u u c c sin cos d u cos sin sin sin cos sin u sin u u cos c c (iii) d u d u u d u. u ln u c c ln( ) (iv) 0 u MEI, 7/0/07 5/7

6 When = 0, u = When =, u = = u d ( u ) u u u ( u u ) For definite integration, the limits of integration must be changed from values of to values of u. Integration by parts This technique comes directly from the proct rule for differentiation, and so is often appropriate for dealing with integrals which are procts, often with another function of, e.g. cos, e, ln. The formula for integration by parts is dv u uv v Usually the simpler function (often ) is taken to be u. However, integrals which are procts including ln are dealt with slightly differently by taking u = ln. Eample 6 Find using integration by parts: cos d ln d Solution cos d Let u dv cos v sin cos sin sin sin sin sin cos c 9 ln d Let u ln MEI, 7/0/07 6/7

7 dv v ln ln ln ln c 9 For some practice in choosing the appropriate technique, try the Mathsnet resource Match them up. MEI, 7/0/07 7/7